Exam 2024 (B)

1a

\begin{matrix} Let x > 0 \\ Prove: \frac{\arctan (\frac{x}{x + 1})}{\ln (x + 1)} > \frac{x + 1}{2 x^{2} + 2 x + 1} \\ Proof: \\ Let f (x) = \arctan (\frac{x}{x + 1}) \\ f^{'} (x) = {(\frac{x}{x + 1})}^{'} \cdot \frac{1}{{(\frac{x}{x + 1})}^{2} + 1} = \frac{x + 1 - x}{(x + 1)^{2}} \cdot \frac{1}{\frac{2 x^{2} + 2 x + 1}{x^{2} + 2 x + 1}} = \frac{1}{2 x^{2} + 2 x + 1} \\ Let g (x) = \ln (x + 1) \\ g^{'} (x) = \frac{1}{x + 1} \\ \frac{f^{'} (x)}{g^{'} (x)} = \frac{x + 1}{2 x^{2} + 2 x + 1} \\ \frac{f (x)}{g (x)} = \frac{f (x) - 0}{g (x) - 0} = \frac{f (x) - f (0)}{g (x) - g (0)} \\ By Langranges theorem: \exists c \in (0, x) : \frac{f (x) - f (0)}{g (x) - g (0)} = \frac{f^{'} (c)}{g^{'} (c)} \\ Let h (x) = \frac{f^{'} (x)}{g^{'} (x)} \\ h^{'} (x) = \frac{2 x^{2} + 2 x + 1 - (x + 1) (4 x + 2)}{(2 x^{2} + 2 x + 1)^{2}} = \frac{- 2 x^{2} - 4 x - 1}{(2 x^{2} + 2 x + 1)^{2}} < 0 \\ ⟹ [c < x ⟹ h (c) > h (x)] \\ ⟹ \frac{f (x)}{g (x)} > h (x) ⟹ \frac{\arctan (\frac{x}{x + 1})}{\ln (x + 1)} > \frac{x + 1}{2 x^{2} + 2 x + 1} \end{matrix}

\begin{matrix} Let f : R \to R be a function \\ Let f be continuous on R \\ Prove or disprove: f (x) = x has a solution ⟺ f (f (x)) = x has a solution \\ Proof: \\ Let \exists c \in R : f (c) = c \\ ⟹ f (f (c)) = f (c) = c \\ Let \exists c \in R : f (f (c)) = c \\ Let g (x) = f (x) - x \\ g is continuous on R \\ g (c) = f (c) - c \\ g (f (c)) = f (f (c)) - f (c) = c - f (c) \\ If f (c) - c = 0 then f (c) = c \\ If f (c) - c \neq 0 ⟹ g (c) \cdot g (f (c)) < 0 \\ ⟹ By the intermediate value theorem: \exists d \in (c, f (c)) : g (d) = 0 \\ ⟹ f (d) = d \end{matrix}

\begin{matrix} Find: lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} \\ Solution: \\ Let t = \frac{1}{n} \\ (e^{t} - e^{- t})^{\sin t} = {(\frac{e^{2 t} - 1}{e^{t}})}^{\sin t} = e^{\ln (\frac{e^{2 t} - 1}{e^{t}}) \sin t} = e^{\ln (e^{2 t - 1}) \sin t - \ln (e^{t}) \sin t} = \\ = e^{\ln (e^{2 t} - 1) \sin t - t \sin t} \\ lim_{t \to 0} \frac{\ln (e^{2 t} - 1)}{\frac{1}{\sin t}} = lim_{t \to 0} \frac{t \ln (e^{2 t} - 1)}{\frac{t}{\sin t}} = lim_{t \to 0} \frac{\ln (e^{2 t} - 1)}{\frac{1}{t}} = \\ \overset{L}{=} lim_{t \to 0} \frac{\frac{2 e^{2 t}}{e^{2 t} - 1}}{- \frac{1}{t^{2}}} = lim_{t \to 0} \frac{2 e^{2 t} t^{2}}{1 - e^{2 t}} \overset{L}{=} lim_{t \to 0} \frac{4 e^{2 t} t^{2} + 4 e^{2 t} t}{1 - \underset{\to 2}{\underset{⏟}{2 e^{2 t}}}} \to \frac{0 + 0}{- 1} = 0 \end{matrix}

\begin{matrix} Find: lim_{n \to \infty} {(1 + \frac{1}{n})}^{(- 1)^{n} \cdot n} \\ Solution: \\ Let a_{n} = {(1 + \frac{1}{n})}^{(- 1)^{n} \cdot n} \\ a_{2 n} = {(1 + \frac{1}{2 n})}^{2 n} \to e \\ a_{2 n - 1} = {(1 + \frac{1}{2 n - 1})}^{- (2 n - 1)} = \frac{1}{{(1 + \frac{1}{2 n - 1})}^{2 n - 1}} \to \frac{1}{e} \\ lim_{n \to \infty} a_{2 n} \neq lim_{n \to \infty} a_{2 n - 1} ⟹ ∄ lim_{n \to \infty} a_{n} \end{matrix}

\begin{matrix} Let a_{n} be monotonically non-increasing \\ Prove or disprove: \sum_{n = 1}^{\infty} a_{n} converges ⟹ n a_{n} \to 0 \\ Proof: \\ Let \sum_{n = 1}^{\infty} a_{n} converges \\ ⟹ \sum_{n = 1}^{\infty} 2^{n} a_{2^{n}} converges \\ ⟹ a_{n} \to 0 and 2^{n} a_{2^{n}} \to 0 \\ \forall n \in N : \exists k \in N : 2^{k} \leq n \leq 2^{k + 1} \\ a_{n} is monotonically non-increasing ⟹ a_{2^{k + 1}} \leq a_{n} \leq a_{2^{k}} \\ ⟹ 2^{k} a_{2^{k + 1}} \leq n a_{n} \leq 2^{k + 1} a_{2^{k}} \\ ⟹ \frac{1}{2} \underset{\to 0}{\underset{⏟}{2^{k + 1} a_{2^{k + 1}}}} \leq n a_{n} \leq 2 \cdot \underset{\to 0}{\underset{⏟}{2^{k} a_{2^{k}}}} \\ ⟹ n a_{n} \to 0 \end{matrix}

\begin{matrix} Let a_{n} be monotonically non-increasing \\ Prove or disprove: n a_{n} \to 0 ⟹ \sum_{n = 1}^{\infty} a_{n} converges \\ Disproof: \\ Let a_{n} = \frac{1}{n \ln n} \\ n a_{n} = \frac{1}{\ln n} \to 0 \\ \sum_{n = 1}^{\infty} 2^{n} a_{2^{n}} = \sum_{n = 1}^{\infty} \frac{2^{n}}{2^{n} n \ln (2)} = \frac{1}{\ln (2)} \sum_{n = 1}^{\infty} \frac{1}{n} diverges \\ ⟹ \sum_{n = 1}^{\infty} a_{n} diverges \end{matrix}

\begin{matrix} Let a_{1}, b_{1} > 0 \in R \\ Let a_{n + 1} = \frac{a_{n} + b_{n}}{2} \\ Let b_{n + 1} = \sqrt{a_{n} b_{n}} \\ Prove: lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} = L \in R \\ Proof: \\ Base case. a_{1} > 0, b_{1} > 0 \\ Induction step. Let a_{n} > 0, b_{n} > 0 \\ a_{n + 1} = \frac{a_{n} + b_{n}}{2} > 0 \\ b_{n + 1} = \sqrt{a_{n} b_{n}} > 0 \\ ⟹ \forall n \in N : a_{n} > 0, b_{n} > 0 \\ \frac{a_{n} + b_{n}}{2} - \sqrt{a_{n} b_{n}} = \frac{a_{n} - 2 \sqrt{a_{n} b_{n}} + b_{n}}{2} = \frac{(\sqrt{a_{n}} - \sqrt{b_{n}})^{2}}{2} > 0 \\ ⟹ a_{n + 1} > b_{n + 1} \\ ⟹ \forall n > 1 \in N : b_{n} < a_{n} \\ ⟹ \forall n > 1 \in N : a_{n + 1} < \frac{2 a_{n}}{2} = a_{n} \\ ⟹ a_{n} is monotonically decreasing and bounded from below by 0 \\ ⟹ \exists lim_{n \to \infty} a_{n} = L \in R \\ b_{n} = 2 a_{n + 1} - a_{n} ⟹ lim_{n \to \infty} b_{n} = lim_{n \to \infty} 2 a_{n + 1} - a_{n} = 2 L - L = L \\ ⟹ lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} = L \in R \end{matrix}

\begin{matrix} Let f be a function defined on (a, b) \\ Let f be differentiable three times on (a, b) \\ Let \exists c \in (a, b) : f^{'} (c) = 0 \\ Prove or disprove: f^{″} (c) > 0 ⟹ c is a local minimum \\ Proof: \\ f^{″} (c) = lim_{h \to 0} \frac{f^{'} (c + h) - f^{'} (c)}{h} = lim_{h \to 0} \frac{f^{'} (c + h)}{h} > 0 \\ h \to 0^{+} ⟹ f^{'} (c + h) > 0 ⟹ \exists ε_{R} > 0 : f is monotonically increasing on [c, c + ε_{E}] \\ h \to 0^{-} ⟹ f^{'} (c + h) < 0 ⟹ \exists ε_{L} > 0 : f is monotonically decreasing on [c - ε_{L}, c] \\ ⟹ \exists ε = m i n (ε_{L}, ε_{R}) > 0 : \forall x \in [c - ε, c + ε] : f (c) \leq f (x) \\ ⟹ c is a local minimum of f \end{matrix}